Askiitians Tutor Team
Last Activity: 5 Months ago
To solve this problem, we need to analyze the situation involving the three cars and their speeds. Car A is moving at a constant speed, while Cars B and C are approaching from opposite directions. The goal is to determine the minimum acceleration required for Car B to safely overtake Car A before Car C reaches Car A. Let’s break this down step by step.
Understanding the Initial Conditions
We have the following details:
- Speed of Car A: 36 km/hr
- Speed of Cars B and C: 54 km/hr each
- Distance from Car A to Car B (AB): 1 km
- Distance from Car A to Car C (AC): 1 km
Converting Speeds to Consistent Units
First, let's convert the speeds from kilometers per hour to meters per second for easier calculations:
- Car A: 36 km/hr = 36 * (1000 m / 3600 s) = 10 m/s
- Car B: 54 km/hr = 54 * (1000 m / 3600 s) = 15 m/s
- Car C: 54 km/hr = 15 m/s
Calculating Time Until Collision
Next, we need to determine how long it will take for Cars B and C to reach Car A if they maintain their current speeds. Since both Cars B and C are 1 km away from Car A, we can calculate the time it takes for each car to reach Car A:
- Time for Car B to reach A: Distance / Speed = 1000 m / 15 m/s = 66.67 seconds
- Time for Car C to reach A: Distance / Speed = 1000 m / 15 m/s = 66.67 seconds
Distance Covered by Car A
During this time, Car A will also be moving. In 66.67 seconds, Car A will cover:
Distance = Speed x Time = 10 m/s x 66.67 s = 666.67 m
Positioning of Cars After 66.67 Seconds
After 66.67 seconds, Car A will be approximately 666.67 m down the road, leaving a distance of:
Remaining distance from Car B to Car A = 1000 m - 666.67 m = 333.33 m
Acceleration Required for Car B
Now, Car B needs to cover this remaining distance of 333.33 m before Car C reaches Car A. We can use the kinematic equation to find the required acceleration:
Using the equation: d = v_i * t + 0.5 * a * t^2, where:
- d = 333.33 m (distance to cover)
- v_i = 15 m/s (initial speed of Car B)
- t = 66.67 s (time until Car C reaches Car A)
Plugging in the values, we have:
333.33 = 15 * 66.67 + 0.5 * a * (66.67)^2
333.33 = 1000.05 + 0.5 * a * 4444.89
Now, rearranging the equation to solve for acceleration (a):
0.5 * a * 4444.89 = 333.33 - 1000.05
0.5 * a * 4444.89 = -666.72
a = (-666.72 * 2) / 4444.89
a ≈ -0.30 m/s²
Conclusion on Acceleration
Since acceleration cannot be negative for Car B to overtake Car A, this indicates that Car B must already be traveling faster than Car A and maintain that speed to avoid an accident. Therefore, Car B does not need to accelerate; it simply needs to maintain its speed of 15 m/s to safely overtake Car A before Car C arrives.
In summary, Car B can safely overtake Car A without any additional acceleration, as long as it maintains its speed of 15 m/s. This ensures that it reaches Car A before Car C does, avoiding any potential collision.
